P7714 "EZEC-10" Sorting Problem Solution

difference array

Every time we input, if $x=i$ （ $x$ is the input number) you can directly$continue$ (because it is the minimum value), if$x!=i$ , save the largest and smallest of the two first, and record the difference array.

Why use differential arrays?

Because we found that if $i$ to$The\; numbers\; in\; j$ are sorted, then the contribution of each number is 1, we use the difference array to save it, and finally calculate the prefix sum. If the difference array of a point is not zero, we can$ans++$ , without considering the specific value of its difference array (because it is the minimum value).

---

$Code$

#include<bits/stdc++.h>
using namespace std;

#define int long long
int t;
const int maxn = 1000005;
int n, a[maxn];

inline int read ()
{
    
    
	int s = 0, x = 1;
	char ch;
	ch = getchar ();
	while (ch < '0' or ch > '9')
	{
    
    
		if (ch == '-') x = -1;
		ch = getchar ();
	} 
	while (ch >= '0' and ch <= '9')
	{
    
    
		s = s * 10 + ch - '0';
		ch = getchar ();
	}
	return s * x;
}

signed main ()
{
    
    
	t = read ();
	while (t--)
	{
    
    
		n = read ();
		for (int i = 1; i <= n; i++) a[i] = 0;
		for (int i = 1; i <= n; i++)
		{
    
    
			int x;
			x = read ();
			if (x == i) continue;
			int maxx = max (x, i);
			int minn = min (x, i);
			a[maxx + 1]--;
			a[minn]++;
		}
		int ans = 0;
		for (int i = 1; i <= n; i++)
		{
    
    
			a[i] += a[i - 1];
			if (a[i]) ans += 1;
		}
		printf ("%lld\n", ans);
	}
	return 0;
}

——$End$——